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Chapter 1: Problem 24

Evaluate \(g(-x), g(2 x),\) and \(g(a+h)\). $$g(x)=\sqrt{5}$$

Short Answer

Expert verified
The results are \(g(-x) = \(\sqrt{5}\), \(g(2x) = \(\sqrt{5}\), and \(g(a+h) = \(\sqrt{5}\).

Step by step solution

01

Evaluate \(g(-x)\)

The function \(g(x)\) is defined to be \(\sqrt{5}\), which is a constant. So, the result of \(g(-x)\) is also \(\sqrt{5}\), regardless of the value of x.
02

Evaluate \(g(2x)\)

Again, because \(g(x)\) is a constant function and equals \(\sqrt{5}\) for any x, the result of \(g(2x)\) is also \(\sqrt{5}\).
03

Evaluate \(g(a+h)\)

Similar to the previous steps, the value of \(g(a+h)\) will also be \(\sqrt{5}\). This is because \(g(x)\) is a constant function and it doesn't change its value based on the input.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a crucial concept in mathematics, particularly when dealing with functions such as the one presented in the exercise. The term refers to the process of substituting a given value or expression into a function to determine its output. In simpler terms, it's about finding out what result the function gives when you "plug in" the input.
For a constant function like \(g(x) = \sqrt{5}\), the evaluation is straightforward.
  • No matter what input (or expression) you substitute, such as \(-x\), \(2x\), or \(a+h\), the output will remain \(\sqrt{5}\).
  • This is because the function is constant, meaning it has the same value for all inputs.
  • Unlike linear or quadratic functions, constant functions do not change based on input values.
Understanding function evaluation helps build a solid foundation for more complex mathematical concepts, ensuring clarity and precision in calculations.
Precalculus
Precalculus serves as a bridge between algebra, geometry, and calculus, providing the necessary groundwork and conceptual knowledge for studying calculus. Constant functions like the one in the exercise fall within the precalculus spectrum. Here's why:
  • They introduce students to the broader category of functions, which is essential in calculus.
  • Understanding how different types of functions behave prepares students for the study of limits and derivatives in calculus.
  • Constant functions, such as \(g(x) = \sqrt{5}\), exemplify functions with no rate of change, a critical precursor to the concept of the derivative, which measures change.
Precalculus aims to develop mathematical thinking, ensuring students are equipped for the challenges presented in higher level math courses by reinforcing a deep understanding of fundamental concepts like function behavior and evaluation.
Algebraic Functions
Algebraic functions comprise one of the broad categories of functions studied in mathematics. They include constant functions, polynomial functions, and rational functions, among others. The current exercise focuses on a constant function \(g(x) = \sqrt{5}\), which can be considered one of the simplest algebraic functions.
Here are some key points about algebraic functions:
  • Algebraic functions are constructed using algebraic operations such as addition, subtraction, multiplication, division, and finding roots.
  • Constant functions, which are a type of algebraic function, always return the same value regardless of input. They can be expressed in the form \(f(x) = c\), where \(c\) is a constant.
  • Understanding the characteristics of algebraic functions aids in comprehending more complex mathematical ideas, including their manipulation and graphing.
Recognizing how constant functions fit into the broader category helps in understanding their role and applications within mathematical problem solving.

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